Question

If $\int \frac{1}{\sqrt[5]{(x - 1)^4}(x + 3)^6} \, dx = A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C,$ where $C$ is the constant of integration, then the value of $\alpha + \beta + 20AB$ is _______.

Answer 1

Consider the given integral:
$I = \int \frac{1}{(x - 1)^{4/5}(x + 3)^{6/5}} dx.$
To simplify, let:
$I = \int \frac{1}{(x - 1)^{4/5}(x + 3)^{6/5}} dx = \int \frac{1}{(x - 1)^{4/5}(x + 3)^2} dx.$
Substitute:
$t = \frac{x - 1}{x + 3} \implies dt = \frac{4}{(x + 3)^2} dx.$
Thus, the integral becomes:
$I = \frac{1}{4} \int t^{-4/5} dt.$
Integrating:
$I = \frac{1}{4} \cdot \frac{t^{1/5}}{1/5} + C = \frac{5}{4} t^{1/5} + C.$
Substituting back $t = \frac{x - 1}{x + 3}$:
$I = \frac{5}{4} \left(\frac{x - 1}{x + 3}\right)^{1/5} + C.$
Comparing with:
$\int \frac{1}{\sqrt[5]{(x - 1)^4(x + 3)^6}} dx = A \left(\frac{\alpha x - 1}{\beta x + 3}\right)^B + C.$
We have:
$A = \frac{5}{4}, \quad \alpha = \beta = 1, \quad B = \frac{1}{5}.$
Calculating $\alpha + \beta + 20AB$:
$\alpha + \beta + 20AB = 1 + 1 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7.$
Answer: 7.